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The conditions for steady current are often specified as $$\frac{\partial\rho}{\partial t}=0 \,\,\,\,and\,\,\,\frac{\partial\vec{J}}{\partial t}=0 $$ Combining $\frac{\partial\rho}{\partial t}=0$ with the continuity equation ($\nabla\cdot \vec{J}=\frac{\partial\rho}{\partial t}$) we get that for a steady current, we must have that $$\nabla \cdot \vec{J}=0$$ That is, the divergence of the current density must be zero everywhere for a steady current. I take issue with this implication for suppose we have an a infinite wire of uniform conductivity $\sigma$ and of cross sectional area $A$ conducting a steady current with current density $\vec{J}$. The current density at every point within the wire is clearly the same. However outside the wire, the current density is zero everywhere (we can assume the wire is immersed in a perfect insulator). That means that at the boundary between the wire and its surroundings, $\vec{J}$ experiences a discontinuous drop. Now my question is whether it is actually correct to say that in steady current conditions, we must necessarily have that $\nabla \cdot \vec{J}=0$. This surely can't be correct because I have just used the most stereotypical and idealized example of steady current (an ideal and infinite wire with truly uniform conductivity) and have shown that even in this extremely simplified and idealized case, we do not have that $\nabla \cdot \vec{J}=0$ for all points in space. So what is going on here? Also, what would the implications of this be for the charge distribution at the boundary? From ohms law, we have that $$\vec{E}=\rho \vec{J}$$ $$\Rightarrow \nabla \cdot \vec{E} =\nabla \cdot (\rho \vec{J})$$ Clearly the RHS of the above is undefined at the boundary (both $\sigma$ and $\vec{J}$) experience discontinuities there. So that means the LHS, namely $\vec{E}$ is also undefined at the boundary. From gauss's law, does this not mean that the charge density at the boundary is undefined?

Any help on these issues would be most appreciated!

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2 Answers 2

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Suppose that we have an infinite wire of radius $a$ on the $z$ axis. An steady current in cylindrical coordinates can be described as

$$\vec{J}=J\Theta(a-\rho)\hat{z},$$

where $\rho$ is the axial distance and $\Theta$ is the Heaviside step function.

If you take the divergence of this current density, it takes this form in cylindrical coordinates

$$\vec{\nabla}\cdot\vec{J}=\frac{1}{\rho}\frac{\partial(\rho J_\rho)}{\partial\rho}+\frac1\rho\frac{\partial J_\phi}{\partial\phi}+\frac{\partial J_z}{\partial z}.$$

The only non-zero component of the current density is $J_z$, but it is independent of $z$. We also have to check what happens to this divergence at $\rho=0$, because the first two terms are undefined in that region. To do this, we could integrate $\vec{\nabla}\cdot(A\hat{z})$ in a infinite cylinder of radius $\varepsilon$ over the $z$ axis

$$\iiint_V\vec{\nabla}\cdot(A\hat{z})\,dV=\iint_SA\hat{z}\cdot d\vec{S}=\iint_SA\hat{z}\cdot\hat{\rho}\,dS=0,$$

so $\vec{\nabla}\cdot\vec{J}=0$ everywhere.

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  • $\begingroup$ Thanks for the excellent answer! Okay so basically my assumption that $\nabla \cdot \vec{J}$ is undefined at the boundary because $\vec{J}$ experiences a discontinuity there is simply an incorrect assumption? That is, the divergence of a vector field at a discontinuous point is not necessarily undefined at that point? $\endgroup$ Mar 13, 2021 at 11:15
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    $\begingroup$ In this case, what suffers a discontinuity is the $J_z$ component, but that discontinuity is in the "radial" direction, and since there is no $\frac{\partial J_z}{\partial \rho}$ derivative, everything works fine. $\endgroup$
    – AFG
    Mar 13, 2021 at 11:41
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My guess would be that a current density function isnt defined outside the wire. all points at which the J function is defined shows no change. div j =0 can be retrieved directly from the field equations when de/dt is 0.

take a pillbox of a straight wire there is no radial component of j outside the wire. therefore the only flux in and out of the box is in the direction of the wire therefore zero net flux and thus zero divergence

edit: even given that it is defined as zero and there IS a dicontinuity. that still doesnt prove div j isnt zero as... as youve stated there is zero j outside wire meaning a pillbox wouldnt show any radial component of j outside the wire

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  • $\begingroup$ However another point to add is that biotsavart law for steady currents... is actually not technically valid. by steady currents I mean that the actual magnitude of current is steady.however if the direction of the current changes e.g in coils there is a changing current density and therefore a changing electric field . (current density = k electric field) and radiation would actually be produced on the wire bends( although small so we ignore it even for steady currents) $\endgroup$ Mar 13, 2021 at 9:00

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